Density of $X_1 + X_2$ using convolution

Question

Let $X_1$ and $X_2$ be independent random variables continuously uniformly distributed on $[−1, 1]$ and $[0, 5]$, respectively.

Determine, with intermediate steps, the density of $X_1 + X_2$. To do this, use the convolution formula.

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I have done the following :

Let $Z=X+Y$. Then \begin{align*}\int f_Z(z)\,dz&=P(\{X+Y=z\})\\ & =P(\{X=t,Y=z−t\})\\ & =P(\{X=t\})\cdot P(\{Y=z−t\})\\ & =\int f_X(t)\, dt\cdot \int f_Y(z-t)\, dt\end{align*}

Is that correct so far ?

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EDIT :

Using the formula of convolution, do we have the following ? \begin{align*}f_{X_1+X_2}&=\int_{\mathbb{R}}f_{X_1}(t)f_{X_2}(z-t)\, dt\\ &=\int_{-\infty}^{-1}f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{1}^{+\infty}f_{X_1}(t)f_{X_2}(z-t)\, dt\\ &=\int_{-\infty}^{-1}0\cdot f_{X_2}(z-t)\, dt+\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{1}^{+\infty}0\cdot f_{X_2}(z-t)\, dt\\ &=\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt\\ &=\int_{-1}^1\frac{1}{2}f_{X_2}(z-t)\, dt\\ &=\frac{1}{2}\cdot \int_{-1}^1f_{X_2}(z-t)\, dt\end{align*} Then we have that $f_{X_2}(z-t)=\frac{1}{5}$ if $z-t\in [0,5]$ and otherwise $0$, or not?

Answer 1

We have $~X \sim U(-1, 1)$ and $Y \sim U(0, 5)$.

For distribution of $Z = X + Y, ~F_Z(z) = P(Y \lt z - x)$

Now as we must have $0 \lt y \lt z - x \lt 5$,

$(i)$ for $- 1 \lt z \lt 1, - 1 \lt x \lt z $

$(ii)$ For $1 \lt z \lt 4, -1 \lt x \lt 1$

$(iii)$ For $4 \lt z \lt 6$,
if $~-1 \lt x \lt z-5, 0 \lt y \lt 5$
if $~z - 5 \lt x \lt 1, 0 \lt y \lt z - x$

Please see the below diagram to understand better. Blue line is the upper bound of $z$ in $(i)$, the red line corresponds to upper bound of $z$ in $(ii)$ and the green line is the upper bound of $z$ in $(iii)$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Hereafter, $\ds{\bracks{\cdots}}$ is an Iverson Bracket. Namely, $\ds{\bracks{P} =1}$ whenever $\ds{P}$ is true and $\ds{0}$ otherwise. \begin{align} &\bbox[5px,#ffd]{\int_{0}^{5}{1 \over 5}{1 \over 2}\bracks{-1 < z - x_{2} < 1}\dd x_{2}} = {1 \over 10}\int_{0}^{5}\bracks{z - 1 < x_{2} < z + 1} \dd x_{2} \\[5mm] = & \ {1 \over 10}\bracks{\verts{z} < 1}\int_{0}^{z + 1}\dd x_{2} + {1 \over 10}\bracks{1 < z < 4}\int_{z - 1}^{z + 1}\dd x_{2} \\[2mm] + & {1 \over 10}\bracks{4 \leq z <6}\int_{z - 1}^{5}\dd x_{2} \\[5mm] = & \ \bbox[5px,#ffd]{{\bracks{\verts{z} < 1}\pars{z + 1} + 2\bracks{1 \leq z < 4} + \bracks{4 \leq z < 6}\pars{6 - z}\over 10}} \end{align} The above expression vanishes out whenever $\ds{z < -1\ or\ z > 6}$.